To be more precise, disjoint intervals define the blocks. If [2,5] establishes a block, then Mi,j could be nonzero when i is between 2 and 5, and j is between 2 and 5. Everything outside the specified blocks is zero. Here is a simple 4x4 block matrix, with blocks [1,2] and [3,4].

7 | 9 | 0 | 0 |

4 | 2 | 0 | 0 |

0 | 0 | 8 | 6 |

0 | 0 | 1 | 1 |

You may have noticed that every matrix is block diagonal, consisting of one block running from 1 to n. Well that's not very interesting, so some books say a block diagonal matrix has to have at least two blocks.

Let the first block, in the upper left, be defined by the interval [1,j]. Use the product permutation definition, which I called det2(M) in the section on determinants. Here the determinant is a sum of products, one product for each permutation on n columns.

If a product is to be nonzero, the first row must select an entry from the first j columns. The second row selects an entry from the first j columns, not repeating the column selected earlier. The third row selects an entry in column [1,j], and so on through the first j rows. This is step 1 in building a permutation.

In step 2 we select columns for rows j+1 through n. The columns selected are all larger than j, else the product falls to 0.

for each possible arrangement in step 1, a permutation on the first j numbers, step 2 can make selections in all possible ways, with row and column numbers beyond j. But this is just the determinant of the lower submatrix, which is equal to the product of the determinants of its blocks (by induction). Multiply this determinant by the sum of the permutation products for the first j rows, which is the determinant of the first block. Put this all together and the determinant of M is the product of the determinants of the individual blocks.

As a check, apply this result to a diagonal matrix, where each block is a single element. Sure enough, the determinant is the product of the elements on the main diagonal.

For an example, reconsider the 4x4 matrix presented earlier. Here it is again, with the permutation products shown below.

7 | 9 | 0 | 0 |

4 | 2 | 0 | 0 |

0 | 0 | 8 | 6 |

0 | 0 | 1 | 1 |

+7×2 × 8×1

-7×2 × 6×1

-9×4 × 8×1

+9×4 × 6×1

+7×2 × det(lower block)

-9×4 × det(lower block)

det(upper block) × det(lower block)

This formula doesn't change if we fill in the upper right with nonzero values, creating an upper triangular block matrix. Show this for two blocks, and it holds for any number of blocks by induction. Suppose a permutation product employs a nonzero value in the upper right block. The rest of the rows in the upper left block can consume all but one of the columns in the upper left block. The remaining column is used by a lower row, where the entries are all 0. These products contribute nothing, and we are back to the product of the determinants of the blocks on the main diagonal.

Similarly, the trace of the matrix is the sum of the traces of the individual blocks.